Question

A simple pendulum completes 20 vibrations
in 43 seconds

FIND IT'S
TIME PERIOD AND LENGTH OF PENDULUM (TAKE G=9.8ms-²)​

Answer 1

Answer

Time taken to complete 20 oscillations =32s

Time taken to complete 1 oscillation =32/20s 1.6s

Time period of a pendulum is time taken by it to complete 1 oscillation.

Time period of pendulum is 1.6 seconds.

Answer 2

Answer:

A simple pendulum completes 20 vibrations  in 43 seconds.

The time period is $2.15\ s$ and length of the pendulum is  $106.63614 m$

Step-by-step explanation:

The time required to complete one oscillation is known as the time period. It is given by,  

$Time\ period\ (T)= \frac{1}{Frequency\ (f)}$                    ...(1)

That is, the inverse of the frequency is the time period of the oscillating body.

The frequency of oscillations of a vibrating body is the ratio of the number of oscillations of that body per second.

It can be expressed as,

$Frequency\ (f) = \frac{Number\ of\ oscillations\ (N)}{Time\ (t)}$     ...(2)

In the question, it is given that,

$N = 20$

$t = 43\ s$

Substituting these values into equation (1) and equation (2)

$f = \frac{20}{43} = 0.465116\ Hz$

Now

$T = \frac{1}{0.465116} =2.15 \ s$

We have the expression connecting length of the pendulum and time period as,

$\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$                 ...(3)    

Given,

$g = 9.8\ m/s^{2}$    $T = 2.15 \ s$

Equation (3) becomes,

$\mathrm{2.15}=2 \pi \sqrt{\frac{l}{9.8}}$  

$\frac{2.15}{2\pi } =\sqrt{\frac{l}{9.8} }$

$(\frac{2.15}{2\pi } )^2 =\frac{l}{9.8} }$

$l = (\frac{2.15}{2\pi } )^2\times 9.8$

  $= 106.63614$ m